How Solve This Task From Russian State Exam?
Introduction
The Russian State Exam is a challenging and competitive exam that requires students to demonstrate their knowledge and skills in various subjects, including mathematics. Algebra and precalculus are essential components of the exam, and students need to be well-prepared to tackle complex problems and equations. In this article, we will focus on solving a specific task from the Russian State Exam, which involves an equation with absolute values and a parameter 'a'. We will break down the solution step by step and provide a detailed explanation of each step.
Understanding the Equation
The given equation is:
(|x− a− 2|+ |x− a+ 2|)2 − a(|x− a− 2|+ |x− a+ 2|) + a2 − 64 = 0
This equation involves absolute values, which can be challenging to work with. However, we can simplify the equation by considering different cases based on the value of 'x' relative to 'a'.
Case 1: x < a - 2
In this case, both (x - a - 2) and (x - a + 2) are negative. Therefore, the equation becomes:
((-x + a + 2) + (-x + a - 2))^2 - a((-x + a + 2) + (-x + a - 2)) + a^2 - 64 = 0
Simplifying the equation, we get:
(-2x + 2a)^2 - a(-2x + 2a) + a^2 - 64 = 0
Expanding and simplifying further, we get:
4x^2 - 8ax + 4a^2 + 2ax - 2a^2 + a^2 - 64 = 0
Combine like terms:
4x^2 - 6ax + 3a^2 - 64 = 0
Case 2: a - 2 ≤ x < a + 2
In this case, (x - a - 2) is negative and (x - a + 2) is positive. Therefore, the equation becomes:
((-x + a + 2) + (x - a + 2))^2 - a((-x + a + 2) + (x - a + 2)) + a^2 - 64 = 0
Simplifying the equation, we get:
(2)^2 - a(2) + a^2 - 64 = 0
Expanding and simplifying further, we get:
4 - 2a + a^2 - 64 = 0
Combine like terms:
a^2 - 2a - 60 = 0
Case 3: x ≥ a + 2
In this case, both (x - a - 2) and (x - a + 2) are positive. Therefore, the equation becomes:
((x - a - 2) + (x - a + 2))^2 - a((x - a - 2) + (x - a + 2)) + a^2 - 64 = 0
Simplifying the equation, we get:
(2x - 2a)^2 - a(2x - 2a) + a^2 - 64 = 0
Expanding and simplifying further, we get:
4x^2 - 8ax + 4a^2 + 2ax -2a^2 + a^2 - 64 = 0
Combine like terms:
4x^2 - 6ax + 3a^2 - 64 = 0
Solving the Equation
We have three cases, and in each case, we have a quadratic equation in 'x'. However, we are interested in finding the values of 'a' that satisfy the equation. To do this, we can set the discriminant of the quadratic equation to be non-negative.
For Case 1, the discriminant is:
(-6a)^2 - 4(4)(3a^2 - 64) ≥ 0
Simplifying the inequality, we get:
36a^2 - 48a^2 + 1024 ≥ 0
Combine like terms:
-12a^2 + 1024 ≥ 0
Divide by -12:
a^2 ≤ 85.33
Take the square root:
|a| ≤ √85.33
For Case 2, the discriminant is:
(-2)^2 - 4(1)(a^2 - 60) ≥ 0
Simplifying the inequality, we get:
4 - 4a^2 + 240 ≥ 0
Combine like terms:
-4a^2 + 244 ≥ 0
Divide by -4:
a^2 ≤ 61
Take the square root:
|a| ≤ √61
For Case 3, the discriminant is the same as Case 1:
(-6a)^2 - 4(4)(3a^2 - 64) ≥ 0
Simplifying the inequality, we get:
36a^2 - 48a^2 + 1024 ≥ 0
Combine like terms:
-12a^2 + 1024 ≥ 0
Divide by -12:
a^2 ≤ 85.33
Take the square root:
|a| ≤ √85.33
Conclusion
We have solved the given equation and found the values of 'a' that satisfy the equation. The values of 'a' are:
|a| ≤ √85.33
|a| ≤ √61
These values of 'a' satisfy the equation for all three cases. Therefore, the final answer is:
The values of 'a' that satisfy the equation are |a| ≤ √85.33 and |a| ≤ √61.
Introduction
In our previous article, we solved the task from the Russian State Exam, which involved an equation with absolute values and a parameter 'a'. We broke down the solution step by step and provided a detailed explanation of each step. In this article, we will answer some frequently asked questions related to the solution.
Q: What is the main concept behind solving the equation?
A: The main concept behind solving the equation is to consider different cases based on the value of 'x' relative to 'a'. This allows us to simplify the equation and solve it for each case.
Q: Why did we need to consider three cases?
A: We needed to consider three cases because the equation involves absolute values, which can be positive or negative depending on the value of 'x' relative to 'a'. By considering three cases, we can cover all possible scenarios and solve the equation for each case.
Q: How did we simplify the equation in each case?
A: In each case, we simplified the equation by expanding and combining like terms. This allowed us to isolate the variable 'x' and solve the equation for each case.
Q: What is the discriminant, and why did we need to consider it?
A: The discriminant is a value that can be calculated from the coefficients of a quadratic equation. In this case, we needed to consider the discriminant to determine the values of 'a' that satisfy the equation. If the discriminant is non-negative, then the equation has real solutions.
Q: What are the values of 'a' that satisfy the equation?
A: The values of 'a' that satisfy the equation are |a| ≤ √85.33 and |a| ≤ √61.
Q: Why are there two different values of 'a' that satisfy the equation?
A: There are two different values of 'a' that satisfy the equation because the equation involves absolute values, which can be positive or negative depending on the value of 'x' relative to 'a'. By considering two different cases, we can cover all possible scenarios and solve the equation for each case.
Q: How can I apply this solution to other problems?
A: You can apply this solution to other problems that involve absolute values and quadratic equations. By considering different cases and simplifying the equation, you can solve the problem and find the values of the variables.
Q: What are some common mistakes to avoid when solving equations with absolute values?
A: Some common mistakes to avoid when solving equations with absolute values include:
- Not considering all possible cases
- Not simplifying the equation correctly
- Not isolating the variable correctly
- Not checking the discriminant to ensure real solutions
Q: How can I practice solving equations with absolute values?
A: You can practice solving equations with absolute values by working on problems that involve absolute values and quadratic equations. You can also try solving problems that involve different types of equations, such as linear or polynomial equations.
Conclusion
In this article, we answered some frequently asked questions related to solving the task from the Russian State Exam, which involved an equation with absolute values and a parameter 'a'. We provided a detailed explanation of each step and covered common mistakes to avoid when solving equations with absolute values By practicing solving equations with absolute values, you can improve your skills and become more confident in solving complex problems.